The option D is expressed as the sum of two cubic expressions.
Further explanation:
Given:
The given expressions are as follows:
A. [tex]-64{x^6}{y^{12}}+125{x^{16}}{y^3}[/tex]
B. [tex]-32{x^6}{y^{12}}+125{x^{16}}{y^3}[/tex]
C. [tex]32{x^6}{y^{12}}+125{x^9}{y^3}[/tex]
D. [tex]64{x^6}{y^{12}}+125{x^9}{y^3}[/tex]
Calculation:
Step 1:
For option A the expression [tex]-64{x^6}{y^{12}}+125{x^{16}}{y^3}[/tex] is simplified as follows:
[tex]\begin{aligned}&-64{x^6}{y^{12}}+125{x^{16}}{y^3}\\&\Leftrightarrow-{\left(4 \right)^3}{\left( {{x^2}} \right)^3}{\left( {{y^4}} \right)^3} + {\left( 5 \right)^3}\left( x \right){\left( x \right)^{15}}{\left( y \right)^3} \\ & \Leftrightarrow {\left( {-4{x^2}{y^4}} \right)^3}+x{\left( 5 \right)^3}{\left( {{x^5}} \right)^3}{\left( y \right)^3}\\ &\Leftrightarrow {\left( {-4{x^2}{y^4}} \right)^3}+x{\left( {5{x^5}y} \right)^3} \\ \end{aligned}[/tex]
Thus, the option A is not sum of two cubic expressions.
Step 2:
For option B the expression [tex]-32{x^6}{y^{12}}+125{x^{16}}{y^3}[/tex] is simplified as follows:
[tex]\begin{aligned}&-32{x^6}{y^{12}} + 125{x^{16}}{y^3} \\ & \Leftrightarrow-\left( 4 \right){\left( 2 \right)^3}{\left( {{x^2}} \right)^3}{\left( {{y^4}} \right)^3} + {\left( 5 \right)^3}\left( x \right){\left( x \right)^{15}}{\left( y \right)^3} \\ & \Leftrightarrow4{\left( {-2{x^2}{y^4}} \right)^3} + x{\left( 5 \right)^3}{\left( {{x^5}} \right)^3}{\left( y \right)^3} \\ &\Leftrightarrow4{\left( {-2{x^2}{y^4}} \right)^3} + x{\left( {5{x^5}y} \right)^3} \\ \end{aligned}[/tex]
Thus, the option B is not sum of two cubic expressions.
Step 3:
For option C the expression [tex]32{x^6}{y^{12}} + 125{x^9}{y^3}[/tex] is simplified as follows:
[tex]\begin{aligned} &32{x^6}{y^{12}}+125{x^9}{y^3} \\ &\Leftrightarrow\left( 4 \right)\left( 8 \right){\left( {{x^2}} \right)^3}{\left( {{y^4}} \right)^3} + {\left( 5 \right)^3}{\left( {{x^3}} \right)^3}{\left( y \right)^3} \\ &\Leftrightarrow \left( 4 \right){\left( 2 \right)^3}{\left( {{x^2}} \right)^3}{\left( {{y^4}} \right)^3} + {\left( 5 \right)^3}{\left( {{x^3}} \right)^3}{\left( y \right)^3} \\ &\Leftrightarrow 4{\left( {2{x^2}{y^4}} \right)^3} + {\left( {5{x^3}y} \right)^3} \\ \end{aligned}[/tex]
Thus, the option C is not sum of two cubic expressions.
Step 4:
For option D the expression [tex]64{x^6}{y^{12}}+125{x^9}{y^3}[/tex] is simplified as follows:
[tex]\begin{aligned} &64{x^6}{y^{12}} + 125{x^9}{y^3} \\ &\Leftrightarrow {\left( 4 \right)^3}{\left( {{x^2}} \right)^3}{\left( {{y^4}} \right)^3} + {\left( 5 \right)^3}{\left( {{x^3}} \right)^3}{\left( y \right)^3} \\ &\Leftrightarrow {\left( {4{x^2}{y^4}} \right)^3} + {\left( {5{x^3}y} \right)^3} \\ \end{aligned}[/tex]
Here, the expression [tex]64{x^6}{y^{12}} + 125{x^9}{y^3}[/tex] is written as the sum of two cubic terms [tex]{\left( {4{x^2}{y^4}} \right)^3}[/tex]and[tex]{\left( {5{x^3}y} \right)^3}[/tex].
Thus, the option D is expressed as the sum of two cubic expressions.
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Answer Details :
Grade: Middle School.
Subject: Mathematics.
Chapter: Linear equation.
Keywords:
Expression, cubes, quadratic, cubic expression, linear equation, zeros, 64x^6y^12, function, substitution, pets, direct substitution, elimination, graph, middle term factorization, fraction.
